GEOSYNTHETICS ENGINEERING: IN THEORY AND PRACTICE

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2 GEOSYNTHETICS ENGINEERING: IN THEORY AND PRACTICE Prof. J. N. Mandal Department of civil engineering, IIT Bombay, Powai, Mumbai , India. Tel

3 Module-5 LECTURE- 23 GEOSYNTHETICS IN PAVEMENTS

4 RECAP of previous lecture.. Stress reduction downwards with depth of granular fill Advantages of unpaved roads Design charts of U.S. forest service (USFS) for unpaved roads U.S. forest service design curves Modified California bearing ratio (CBR) test Design of pavement thickness without geogrid (IRC37)

5 Design of Reinforced Unpaved road (Giroud and Han, 2004) Step 1: Calculate the radius of the equivalent contact area using following equation. r P p r = radius of equivalent tire contact area (m), P = wheel load in kn, and p = tire contact pressure; (kpa).

6 Step 2: Calculate the California Bearing Ration of the sub-grade soil or undrained cohesion of sub-grade soil, c u f c CBR sg c u = undrained cohesion of sub-grade soil f C = factor equal to 30kPa C.B.R sg = sub-grade California bearing ratio (< 5) Step 3: Calculate the allowable bearing capacity of sub-grade soil without reinforcement. P h0,unreinforced s fs r 2 N c c u s fs r 2 N f c c CBR sg

7 s = Allowable rut depth is equal to 40 mm, f S = factor equal to 75 mm, r = radius of the equivalent tire contact area, N c = bearing capacity factor equal to 3.14, c u = undrained cohesion of subgrade soil, and CBR sg = subgrade California bearing ratio If the wheel load is greater than the allowable bearing capacity of subgrade soil, a base course is required without or with geosynthetics.

8 To calculate the base course thickness, the following steps should be followed. Step 4: Determine the limit modulus ratio (R E ) and modulus ratio factor (f E ). R E min E E bc sg, 5.0 min 3.48(CBR bc ) 0.3, 5.0 E bc = base course resilient modulus, E sg = subgrade resilient modulus, CBR bc = base course California bearing ratio, and CBR sg = subgrade California bearing ratio, Modulus ratio factor, f E = (R E - 1)

9 Step 5: Determine the bearing capacity mobilization coefficient (m) in term of thickness (h). m s ( ) exp f s r h 2 m = bearing capacity mobilization coefficient, s = allowable rut depth (mm) 50 < s < 100 (as per Giroud, 2004) f S = factor equal to 75 mm r = radius of equivalent tire contact area (m), and h = thickness of base course.

10 Step 6: Determine the require base course thickness h 2 r ( J )( ) log N 2 h P / r 1 r fe mnc fccbrsg J = aperture stability modulus of geogrid (mn/ ) (< 0.8), J = 0 for geotextile and in unreinforced case h = thickness of required base course (m), P = wheel load (kn), and N c = bearing capacity factor = 5.71 (for geogrid) N c, geotextile = 5.14, N c, unreinforced = 3.14 Calculation of the base course thickness needs iterations.

11 If the calculated h value does not match the assumed h value, m value is to be recalculated and is used as the assumed value for the next iteration. The process is to be repeated until the calculated value is approximately equal to assumed value. Step 7: Determine the reduction in base course thickness using geogrid or geotextile (h) = h 0 h Let, h 0 = thickness of required base course for unreinforced case (m) h = thickness of required base course with geogrid or geotextile (m)

12 Example 4.4 Two lane carriageway roads, plain terrain Number of commercial vehicles as per last count (P) =1000 Axle load = 81.6 kn; Wheel load = 81.6/2 = 40.8 kn Tire pressure (p) = 500 kpa Subgrade CBR = 2% Traffic loading category = E [450 to 1500 CVD] Number of years between last count and year of completion of construction (x) = 5 years Annual growth rate of commercial vehicles (r) = 10% Design life of pavement after completion = 15 years

13 Solution: Design as per IRC-37, 2001 (only for unreinforced condition): A P (1 r ) x D N s 365 A [(1 r r ) n 1] x F D = lane distribution factor = 75% (see Table) F = vehicle damage factor = 3.5 (see Table) A 1000 (1 0.1) N s x N s 50 x 10 6 = 50 million standard axles (msa)

14 From design chart, for 2% CBR and 50 msa standard axle, Total pavement thickness = 925 mm = m Design Chart for determining pavement thickness, traffic msa (IRC-37, 2001)

15 Pavement composition (Reference from IRC 37): Bituminous surfacing = 21.5 cm [Bituminous Concrete (B.C.) = 4 cm Bituminous Macadam (D.B.M.) = 17.5 cm] Cross- section of Pavement

16 Design as per Giroud and Han, 2004 (without and with reinforcement): Input data Values Axle load (kn), Q 135 Wheel load (kn), P 67.5 Tire pressure (kpa), p 500 Radius of the equivalent tire contact area (m), r Base course C.B.R. (%), C.B.R. bc 20 C.B.R. of the subgrade soil (%), C.B.R. sg 2 Allowable rut depth (mm), s 40 Factor (kpa), f c 30 Factor (kpa), f s 75 Geogrid aperture stability modulus (m N/ o ), J 0.65 Number of axle passes, N 50 x 10 6 Bearing capacity factor, N c, unreinforced 3.14 N c, geogrid 5.71 N c, geotextile 5.14

17 Output User Interface For unpaved road in unreinforced condition: Assumed base course thickness, h assumed (m) FINISHED #VALUE! Limited modulus ratio, R E 2. Modulus ratio factor, f E Bearing capacity mobilization factor, m #VALUE! #VALUE! Required base course thickness, h final (m) #VALUE! #VALUE!

18 For unpaved road with geogrid: Assumed base course thickness, h assumed (m) FINISHED #VALUE! 1. Limited modulus ratio, R E Modulus ratio factor, f E Bearing capacity mobilization factor, m #VALUE! #VALUE! Required base course thickness, h final (m) #VALUE! #VALUE!

19 For unpaved road with geotextile: Assumed base course thickness, h assumed (m) FINISHED #VALUE! 1. Limited modulus ratio, R E Modulus ratio factor, f E Bearing capacity mobilization factor, m #VALUE! #VALUE! Required base course thickness, h final (m) #VALUE! #VALUE!

20 For geogrid reinforced unpaved road, a factor of safety is considered = 1.5 Percentage of saving (%): Thickness of base course for unreinforced unpaved road (mm), h unreinforced 870 mm Thickness of base course with geogrid (mm), h geogrid 370 x 1.5 = 555 mm Thickness of base course with geotextile (mm), h geotextile 670 mm % Saving (geogrid) 36.21% % Saving (geotextile) 23%

21 Geogrid and geocomposite layers are to be provided in between subgrade and sub-base layers. The geocomposite layer will drain out the infiltrated water from base course and the uplifted water from subgrade. Base course thickness using geogrid = 55.5 cm 56 cm Pavement composition (Geogrid reinforced): Bituminous surfacing (Reference to IRC: 37) = 21.5 cm surfacing consisting of 4 cm bituminous concrete (B.C.) and 17.5 cm dense bituminous macadam (D.B.M.) Base course: 20 cm Water Bound Macadam (W.B.M) (20 % C.B.R.) Sub-base: 35 cm granular material of CBR not less than 30%

22 Unreinforced Reinforced with Geogrid Pavement Section [Traffic (N) = 50 msa, Axle load = 135 kn]

23 Pavement drainage system

24 At 250 mm geogrid can be introduced either at the bottom or in the middle of the base course for specific geogrid Equivalent reinforced base thickness (mm) Geogrid at midpoint of base layer Geogrid at bottom of base layer Nonreinforced base thickness (mm) Geogrid reinforced base course for paved highway (Carroll et al. 1987)

25 AASHTO (1993) Design Method for Flexible Pavement Design problem: A flexible pavement for a road (2 lanes each direction) needs to be designed for 25-years life. Average sub-grade CBR = 2% Expected traffic: tandem axle 36 kips, 100,000/year. No annual growth rate to be considered. Terminal serviceability, p t = 2.5 Reliability level (R) = 95%; standard deviation, S o = 0.35 Asphalt layer coefficient (a 1 ) = 0.40, base course layer coefficient, (a 2 ) = 0.14, sub-base layer coefficient (a 3 ) = Drainage condition is Good ; pavement is exposed to saturation moisture more than 25% of the time.

26 Solution: Without reinforcement: Step 1: Determination of W 18 W 18 = predicted number of 18,000 lb equivalent single axle load (ESAL) applications (a) For tandem axle 36 kips, p t =2.5 and assumed SN of 6.0, the axle load equivalency factor = 1.38 (From Table I) Hence, first year traffic estimate = Expected traffic x axle load equivalency factor = 100,000 x 1.38 = 138,000

27 (b) For analysis period of 25 years and no growth, traffic growth factor = 25 (From Table II) Hence, w 18 = Traffic growth factor x First year traffic estimate = 25 x 138,000 = 3,450,000

28 Table I Axle load equivalency factors for flexible pavements, tandem axles and p t = 2.5 (AASTHO, 1993) Axle load (kips) Axle load equivalency factor Pavement Structural Number (SN)

29 Table II Traffic growth factor (AASTHO, 1993) Analysis period (years) No growth Traffic growth factor Annual Growth Rate, Percent ,

30 (c) Generally for most roadways, D D = 0.5 D D = a directional distribution factor From Table III, D L = 0.90 (two lanes in each direction) Hence, w 18 = D D x D L x w 18 = 0.50 x 0.90 x 3,450,000 = 1,552,500

31 (d) Table IV helps to choose the reliability level. From Table V, Z R = for R = 0.95 Therefore, F R = 10 (-Z R x S o) =10 ( x 0.35) = Finally, W 18 = w 18 x F R = 1,552,500 x = 5,843, msa

32 Table III Lane Distribution Factor, D L (AASTHO, 1993) Number of Lanes in Each Direction Percent of ESAL in Design Lane (D L )

33 Table IV Suggested Levels of Reliability (AASTHO, 1993) Functional Classification Interstate and Other Freeways Recommended Level of Reliability (R) Urban Rural Principal Arterials Collector Local

34 Table V Standard normal deviate for different values of reliability, R Reliability (%) Z R

35 Step 2: Determination of serviceability Given p t =2.5; p o = 4.2 (if not given) p o = Initial design serviceability index PSI = p o - p t = = 1.7 Step 3: Determination of subgrade Resilient Modulus (M R ) M R (psi) = 1,500 x CBR = 1,500 x 2 = 3000 psi M R (kpa) = 6.89 x 1500 x CBR = 6.89 x 1500 x 2 = kpa

36 Step 4: Determination of Structural Number (SN) Basic empirical equation for design of flexible pavement by American Association of State Highway and Transportation (AASHTO, 1993) is as follows (in FPS units). PSI log log (W ) Z.S 9.36 log (SN 1) R o log10(m R ) SN 1 Here, W 18 = , Z R = , S o = 0.35, PSI = 1.7 and M R = 3000 psi Determine the structural number (SN) by trial and error method or use the design charts given. Therefore, SN = 6.16

37 Design chart of CBR vs. SN for 1 to 10 msa (R = 95 % and S o = 0.35)

38 Step 5: Estimation of pavement thickness For unreinforced case, SN = a 1.D 1 + a 2.D 2.m 2 + a 3.D 3.m 3 D 1, D 2 and D 3 = thickness (in inches) of the surface, base and sub-base respectively a 1, a 2, a 3 = structural layer coefficients for the surface, base and sub-base respectively (The layer coefficients depend upon the resilient modulus of the material.) m 2, m 3 = drainage coefficient for base course and sub-base course respectively As SN value has already been determined in step 4, D 1, D 2 and D 3 can be obtained by trial and error.

39 (a) Generally, the typical layer coefficients used by AASTHO road test are as follows: a 1 = Asphalt concrete surface course = , a 2 = Crushed stone base course = , and a 3 = Sandy graved sub-base = We obtain a 1 = 0.40, a 2 = 0.14, a 3 = 0.08 (given) (b) Drainage: Modified layer coefficients are used to account for the improved drainage conditions. The quality of drainage and recommended drainage coefficients can be obtained from Tables VI and VII respectively. As the drainage condition is Good and pavement is exposed to saturation moisture more than 25% of the time, m 2 = 1.0 and m 3 = 1.0

40 Table VI Drainage Conditions (AASTHO, 1993) Quality of Drainage Excellent Good Fair Poor Very poor Duration of Water Removal 2 hours 1 day 1 week 1 month Water will not drain

41 Table VII Recommended m i Values (as per AASTHO, 1993) Quality of Drainage Recommended m i Value Percent of Time Pavement is Exposed to Moisture Levels Approaching Saturation <1% 1-5% 2-25% >25% Excellent Good Fair Poor Very Poor

42 We have already calculated by trial and error, SN = = 0.40.D D D 3.1 or, 6.16 = 0.40 D D D 3 Assuming surface layer depth (D 1 ) = 0.21 m Assuming base course depth (D 2 ) = 0.25 m Hence, sub-base depth (D 3 ) = 0.47 m Total thickness = ( ) = 0.93 m

43 Step 6: Design for Reinforced Case For reinforced case we know that, SN = a 1.D 1 + a 2.D 2.m 2 + (LCR).a 3.D 3.m 3 (N.B. D 1, D 2 and D 3 are in inches) a 1 = 0.40, a 2 = 0.14, a 3 = 0.08 (given) m 2 = 1.0 and m 3 = 1.0 LCR value depends on CBR of subgrade and types of reinforcement used.

44 Here, LCR = 1.4 for CBR = 2

45 Already calculated, SN = = 0.40 x D x D 2 x x 0.08 x D 3 x 1 or, 6.16 = 0.40 D D D 3 Assuming surface Layer Depth (D 1 ) = 0.21 m Assuming base course Depth (D 2 ) = 0.25 m Hence, sub-base Course Depth (D 3 ) = 0.33 m Total thickness = ( ) = 0.79 m

46 Step 7: Saving of sub-base course material (%) unreinforced subbasedepth - reinforcedsubbasedepth % Saving 100 unreinforced subbasedepth % Saving %

47 Pavement Design without and with Geosynthetics (W 18 = 5.8 msa)

48 Please let us hear from you Any question?

49 Prof. J. N. Mandal Department of civil engineering, IIT Bombay, Powai, Mumbai , India. Tel